Answer:
[tex] r= 2.5 m tan(48.61) = 2.84m[/tex]
Explanation:
For this case the situation is explained in the figure attached.
We can find the maximum radius r with the angle of incidence or critical angle (xi). If we have that the angle is higher than the light we will see a reflecting property. We can assume that the refracted angle for this case is 90 and if we use the following trigonometric property we have:
[tex] tan (x_i) = \frac{r}{d}[/tex]
And if we solve for r we got:
[tex] r= d tan(x_i)[/tex]
The value of d on this case represent the depth of the watr d = 2.5 m
And we can find the value for the angle [tex] x_i[/tex] using the Snell's Law like this:
[tex] n_{air} sin (90) = n_{water} x_i [/tex]
[tex] sin (x_i)= \frac{n_{air} sin (90)}{n_{water}}= \frac{1*1}{1.333}= 0.75[/tex]
And we can use the arcsin in order to find the angle like this:
[tex] x_i = arcsin(0.75) =48.61[/tex]
And now we can find the maximum radius replacing into
[tex] r= d tan(x_i)[/tex]
And we got:
[tex] r= 2.5 m tan(48.61) = 2.84m[/tex]
